Answer:
1. 10.7 years
2. 17.0 years
3. 2nd option
Step-by-step explanation:
Use formula for compounded interest
[tex]A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},[/tex]
where
A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.
In your case,
1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then
[tex]11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.[/tex]
2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then
[tex]17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},[/tex]
[tex]3=(1.0054)^{12t},[/tex]
[tex]12t=\log_{1.0054}3,[/tex]
[tex]t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.[/tex]
3. 1 choice: P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then
[tex]A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \$7936.39.[/tex]
2 choice: P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then
[tex]A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.[/tex]
The best will be 2nd option, because $8032.58>$7936.39